Perform the row operation, $4R_1\rightarrow R_1$, on the following matrix. $\left[\begin{array} {ccc} -1 & 8 & 8 & 2 \\ 7 & 5 & 2 & 2 \\ -6 & 7 & 7 & 5 \end{array} \right] $
Solution: Background There are three basic row operations that can be performed on matrices. $R_i \leftrightarrow R_j$. This symbol tells us to interchange rows $i$ and $j$. $cR_i \rightarrow R_i$. This symbol tells us to multiply a row $i$ by a constant $c$. $R_i + cR_j \rightarrow R_i$. This symbol tells us to add $c$ times row $j$ to row $i$. Finding the new row to be used For the given matrix, $R_1$ is given below. $R_1=\left[\begin{array} {ccc} -1 & 8 & 8 & 2 \end{array} \right]$ We are asked to perform the row operation, $4R_1\rightarrow R_1$. Therefore, we must multiply $R_1$ by $4$. $\begin{aligned}4R_1 &= 4\left[\begin{array} {ccc} -1 & 8 & 8 & 2 \end{array} \right] \\\\&=\left[\begin{array} {ccc} -4 &32 & 32 & 8 \end{array} \right]\end{aligned}$ Substituting the row Now, we must substitute row $R_1$ with $4R_1$. $\left[\begin{array} {ccc} {-1} & {8} & {8} & {2} \\ 7 & 5 & 2 & 2 \\ -6 & 7 & 7 & 5 \end{array} \right] \xrightarrow{4R_1\rightarrow R_1} \left[\begin{array} {ccc} {-4} & {32} & {32} & {8} \\ 7 & 5 & 2 & 2 \\ -6 & 7 & 7 & 5 \end{array} \right] $ Summary Our resultant matrix is the following. $\left[\begin{array} {ccc} -4 &32 & 32 & 8 \\ 7 & 5 & 2 & 2 \\ -6 & 7 & 7 & 5 \end{array} \right]$